//Given the root of a binary tree, return the inorder traversal of its nodes' 
//values. 
//
// 
// Example 1: 
//
// 
//Input: root = [1,null,2,3]
//Output: [1,3,2]
// 
//
// Example 2: 
//
// 
//Input: root = []
//Output: []
// 
//
// Example 3: 
//
// 
//Input: root = [1]
//Output: [1]
// 
//
// 
// Constraints: 
//
// 
// The number of nodes in the tree is in the range [0, 100]. 
// -100 <= Node.val <= 100 
// 
//
// 
//Follow up: Recursive solution is trivial, could you do it iteratively? 
//Related Topics Stack Tree Depth-First Search Binary Tree 👍 6418 👎 273

  
package leetcode.editor.en;

import java.util.ArrayList;
import java.util.List;

public class _94_BinaryTreeInorderTraversal{
    public static void main(String[] args) {
         Solution solution = new _94_BinaryTreeInorderTraversal().new Solution();
    }
    //leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> r = new ArrayList<>();
        inorder(root, r);
        return r;
    }

    private void inorder(TreeNode root, List<Integer> r) {
        if (root ==null) {
            return ;
        }
        inorder(root.left, r);
        r.add(root.val);
        inorder(root.right, r);
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}